Explanation:
Since {v1,...,vp} is linearly dependent, there exist scalars a1,...,ap, with not all of them being 0 such that a1v1+a2v2+...+apvp = 0. Using the linearity of T we have that
a1*T(v1)+a2*T(v1) + ... + ap*T(vp) = T(a1v19+T(a2v2)+...+T(avp) = T(a1v1+a2v2+...+apvp) = T(0) = 0.
Since at least one ai is different from 0, we obtain a non trivial linear combination that eliminates T(v1) , ..., T(vp). That proves that {T(v1) , ..., T(vp)} is a linearly dependent set of W.
Given:
Sample Mean <span>= 30<span>
Sample size </span><span><span><span>= 1000</span></span><span>
</span></span></span>Population Standard deviation or <span><span><span>σ<span>=2</span></span><span>
</span></span>Confidence interval </span><span>= 95%</span>
to compute for the confidence interval
Population Mean or <span>μ<span><span>= sample mean ± (</span>z×<span>SE</span>)</span></span>
<span><span>where:</span></span>
<span><span>SE</span>→</span> Standard Error
<span><span>SE</span>=<span>σ<span>√n</span>= 30</span></span>√1000=0.9486
Critical Value of z for 95% confidence interval <span>=1.96</span>
<span>μ<span>=30±<span>(1.96×0.9486)</span></span><span>
</span></span><span>μ<span>=30±1.8594</span></span>
Upper Limit
<span>μ <span>= 30 + 1.8594 = 31.8594</span></span>
Lower Limit
<span>μ <span>= 30 − 1.8594 = <span>28.1406</span></span></span>
<span><span><span>
</span></span></span>
<span><span><span>answer: 28.1406<u<31.8594</span></span></span>
Answer:
Step-by-step explanation:
As the statement is ‘‘if and only if’’ we need to prove two implications
is surjective implies there exists a function 
such that 
.- If there exists a function such that , then is surjective
Let us start by the first implication.
Our hypothesis is that the function is surjective. From this we know that for every 
there exist, at least, one 
such that 
.
Now, define the sets 
. Notice that the set 
is the pre-image of the element 
. Also, from the fact that 
is a function we deduce that 
, and because the sets are no empty.
From each set choose only one element 
, and notice that 
.
So, we can define the function 
as 
. It is no difficult to conclude that 
. With this we have that 
, and the prove is complete.
Now, let us prove the second implication.
We have that there exists a function such that .
Take an element , then 
. Now, write 
and notice that . Also, with this we have that 
.
So, for every element we have found that an element (recall that ) such that , which is equivalent to the fact that is surjective. Therefore, the prove is complete.
